<p>clouder-</p>
<p>You work backwards. You get that f(1) is equal to 2 and that f(3) is equal to 8. 8-2 = 6.</p>
<p>f(2) = 4.
f(x) = f(x+1)/2
f(2) also = f(3)/2.
f(3) = 8
f(1) = f(2)/2 = 2.
8-2 = 6</p>
<p>aah i understand the less than 50 now 
how much do u guys think 2 omit and…uhm 5 wrong would be?</p>
<p>yeah i accidentally omited five since she called time on me before i could go back what do you think my score is gonna be i think i got 2 others wrong as well</p>
<p>@clouder, for f(3)-f(1)
and f(2)=4,(given)
and the equation was
f(n)=f(n+1)/2 (given)</p>
<p>so that means,
f(2)=(f(3)/2),
so 4=f(3)/2, and so f(3) = 8.
then you can solve for f(1)
which would be
f(1)=f(2)/2
so f(1) = 4/2
f(1)=2</p>
<p>then f(3)-f(1) = 8-2, which is 6</p>
<p>what was that one where they gave a series, like, (1 - 1/2), etc., and then (1/n-1 - 1/n)? i didn’t get how the answers corresponded to the given information!</p>
<p>^^lol@ 8-2 being 3.</p>
<p>@rocknpiano Hahahaa whoa! i meant 6.</p>
<p>is everybody sure that x=3 in the 1/g(x) one. i got that answer too, but my reasoning was sort of flawed i think… wouldn’t that make the whole equation undefined not 0?</p>
<p>@ drift 1+ (1-1/2) + (1/2-1/3) + …+ (1/(n-1) - 1/n)
i wrote it out so it is like this:
1+
1-1/2 +
1/2-1/3 +… +
1/(n-2) - 1/(n-1)+
1/(n-1) - 1/n. Clearly all the negative terms cancel with positive ones. leaving
1+1-1/n. Answer: 2-1/n</p>
<p>ok, im going nuts. For the coordinate problem (the 2root13) was one of the answers (8,11)? B/c i don’t recognize (8,12), but I know i did the problem correctly. I also remember that originally I had the y-coordinate off by 1, and then changed.</p>
<p>thanks xD
darn my life</p>
<p>yes what was the series one with the 1, 1/2, (1/2-1/3)…(1/n-1 - 1/n)</p>
<p>@ basketball.
No. because f(x) = 1/g(x). at f(3), 1/g(x) = infinity.
g(3) = 0. as g(3) approaches 0, the function 1/g(x) approaches infinity</p>
<p>driftwood-</p>
<p>Or you can plug in values for n. I tried 2 and 3 for n. The only that worked was 2-(1/n).</p>
<p>really? i got 1 - (1/n) because i used a calc program to find the infinite sum…and the only one that didnt result in a negative sum was the 1-(1/n)</p>
<p>think if that x=3 thing like this:
for g(x) at x=3, g(x) becomes undefined because the value of g(x) (or y or whatever) goes to infinity, as was shown by the asymptote in the graph.
now, if you plug that into the 1/g(x) to find f(x), you get 1/infinity at x=3.
1/infinity is 1 being divided by a ginormous number, which means that the value gets smaller and smaller and smaller, and thus approaches 0, which the question was asking for.</p>
<p>And talking of limits, I also remember that other question which asked for the value that x^x approaches as x approaches 0, and I got 1. I just plugged in 0.00000000001 for x and got 0.0000000001^0.0000000001 is like .999999998 or something. Confirmation?</p>
<p>What was the one that had a function and it asked if the output was divisible by 2, 3, and 4?</p>
<p>@RocknPiano 2 and 3</p>
<p>For the divisibility problem, I just tried out a ton of combinations and got that 2 and 3 always work, but not 4.
So that’s what I put.</p>